 ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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# Optimization Problems: Applications to Economics - Problem 3

Norm Prokup ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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To find the minimum the average cost per unit, first recall that the average cost function is c(x)/x. The domain of this function will be at all positive values of x. Then, find the critical points of the average cost function by finding the zeros of the first derivative. It is possible to find which critical point is a minimum by using the second derivative test (recall that a critical point is a minimum when its second derivative is positive).

Let’s do one more optimization problem involving economics. Find the number x of units that minimizes the average cost per unit which is called c bar if the cost function is c(x) equals 0.004x³ plus 20x plus 1000. Now average cost, this quantity c bar, is defined as the total cost c(x) divided by the number of units produced, x. All you have to do is take this function and divide each term by x. That gives me 0.004x² plus 20 plus 1000x to the -1. This is the average cost function that I have to minimize. I just need to find the value of x that minimizes it.

First thing I like to do is find the feasible domain. Now it makes sense to restrict x to be greater than or equal to zero. It doesn’t make sense to produce negative numbers of items. However in this function, it’s not defined if x equals zero, so I may have to keep x positive. So the feasible domain is going to have to be x greater than zero.

The second thing I do, is find critical points. And for that I need to find the derivative of this function. So c bar prime, the derivative is going to be 0.004 times 2x. That’s 0.008x. The derivative of the plus 20 is zero, and then derivative of the 1000x to the -1. What’s the derivative of x to the -1? It’s -1x to the -2. So this is minus 1000x to the -2.

Now where is this undefined? It’s really only undefined for x equal zero but that’s not in my domain so I don’t have to worry about that. Where does it equal zero? Let’s set this equal to zero and find out. 0.008x equals 1000x to the -2. When I multiply both sides by x² to get rid of this x to the -2, so x², and then I’ll get 0.008x³ equals 1000. And then I could divide both sides by 0.008. Let me do this up here. I get x³ equals 1000 divide by 0.008.

Let me do this division. I’m actually going to do it by multiplying the top and bottom by 1000 to get rid of the decimal. So I have 1000000 over 8. 1000000 over 4 would be 250000, 100000 over 8 is 125000. So this is 125000. That’s x³. X is going to be, well, 125 is 5³ 1000 is 10³ so x is going to be 50.

Now that I have the critical point, what do I do with it? Because my derivative function c' was pretty easy. What was it again? 0.008x minus 1000x to the -1. Because this function is so easy to deal with, I’m going to take the second derivative and use the second derivative test.

C bar double prime is going to be 0.008 minus, the derivative of x to the -2 is -2x to the -3. -2 times -1000 is plus 2000x to the -3. So this is the second derivative. All I have to do to determine whether this number represents an absolute max or an absolute min is to determine whether or not the second derivative is positive or negative over the entire domain. What’s the domain? X greater than zero.

Well for x greater than zero, this quantity is always positive and therefore the whole second derivative is always positive. The second derivative being always positive means that it does this kind of a thing. So wherever there is a critical point, it has an absolute minimum. So c bar has an absolute minimum at x equals 50.

Let’s double check, is this what we were supposed to find? Were we supposed to find the x value that minimizes c bar? So let’s take a look. Find a number x of units that minimizes the average costs per unit c bar, that’s exactly what we’ve found. So it’s true, c bar has an absolute minimum value at x equals 50. We don’t need to find that value.